Advanced Calculus Single Variable

2.8 Arithmetic of Integers

Here we consider some very important algebraic notions including the Euclidean algorithm
just mentioned and issues related to whether two numbers are relatively prime, prime
numbers and so forth. The following definition describes what is meant by a prime number
and also what is meant by the word “divides”.

Definition 2.8.1The number, a divides the number, b if in Theorem 2.7.11,r = 0. That is there is zero remainder. The notation for this is a|b, read a divides b anda is called a factor of b. A prime numberis one which has the property that the onlynumbers which divide it are itself and 1 and it is at least 2. The greatest common divisorof two positive integers, m,n is that number, p which has the property that p dividesboth m and n and also if q divides both m and n, then q divides p. Two integers arerelatively primeif their greatest common divisor is one. The greatest common divisorof m and n is denoted as

(m,n )

.

There is a phenomenal and amazing theorem which relates the greatest common divisor to
the smallest number in a certain set. Suppose m,n are two positive integers. Then if x,y are
integers, so is xm + yn. Consider all integers which are of this form. Some are positive such as
1m + 1n and some are not. The set S in the following theorem consists of exactly
those integers of this form which are positive. Then the greatest common divisor of
m and n will be the smallest number in S. This is what the following theorem
says.

Theorem 2.8.2Let m,n be two positive integers and define

S ≡ {xm + yn ∈ ℕ : x,y ∈ ℤ } .

Then the smallest number in S is thegreatest common divisor, denoted by

(m,n )

.

Proof:First note that both m and n are in S so it is a nonempty set of positive integers.
By well ordering, there is a smallest element of S, called p = x0m + y0n. Either p divides m or
it does not. If p does not divide m, then by Theorem 2.7.11,

m = pq + r

where 0 < r < p. Thus m =

(x0m + y0n)

q + r and so, solving for r,

r = m (1 − x )+ (− y q)n ∈ S.
0 0

However, this is a contradiction because p was the smallest element of S. Thus p|m. Similarly
p|n.

Now suppose q divides both m and n. Then m = qx and n = qy for integers, x and y.
Therefore,

p = mx0 + ny0 = x0qx + y0qy = q(x0x+ y0y)

showing q|p. Therefore, p =

(m,n)

. ■

This amazing theorem will now be used to prove a fundamental property of prime
numbers which leads to the fundamental theorem of arithmetic, the major theorem which
says every integer can be factored as a product of primes.

Theorem 2.8.3If p is a prime and p|ab then either p|a or p|b.

Proof: Suppose p does not divide a. Then since p is prime, the only factors of p
are 1 and p so follows

(p,a)

= 1 and therefore, there exists integers, x and y such
that

1 = ax + yp.

Multiplying this equation by b yields

b = abx + ybp.

Since p|ab, ab = pz for some integer z. Therefore,

b = abx + ybp = pzx+ ybp = p(xz + yb)

and this shows p divides b. ■

Theorem 2.8.4(Fundamental theorem of arithmetic) Let a ∈ℕ∖

{1}

. Thena = ∏i=1npiwhere piare all prime numbers. Furthermore, this prime factorization isunique except for the order of the factors.

Proof:If a equals a prime number, the prime factorization clearly exists. In particular
the prime factorization exists for the prime number 2. Assume this theorem is true for all
a ≤ n − 1. If n is a prime, then it has a prime factorization. On the other hand, if n is not a
prime, then there exist two integers k and m such that n = km where each of k and m are
less than n. Therefore, each of these is no larger than n − 1 and consequently, each has a
prime factorization. Thus so does n. It remains to argue the prime factorization is unique
except for order of the factors.

Suppose

∏n ∏m
pi = qj
i=1 j=1

where the pi and qj are all prime, there is no way to reorder the qk such that m = n and
pi = qi for all i, and n + m is the smallest positive integer such that this happens. Then by
Theorem 2.8.3, p1|qj for some j. Since these are prime numbers this requires p1 = qj.
Reordering if necessary it can be assumed that qj = q1. Then dividing both sides by
p1 = q1,

n−∏ 1 m∏−1
pi+1 = qj+1.
i=1 j=1

Since n + m was as small as possible for the theorem to fail, it follows that n− 1 = m− 1 and
the prime numbers, q2,

⋅⋅⋅

,qm can be reordered in such a way that pk = qk for all
k = 2,

⋅⋅⋅

,n. Hence pi = qi for all i because it was already argued that p1 = q1, and this
results in a contradiction. ■